Question

(\frac{x^2 + 4x + 3}{2x + 1} < 0)

Explanation:

Step1: Factor the numerator

First, factor the quadratic expression in the numerator: \(x^2 + 4x + 3=(x + 1)(x + 3)\). So the inequality becomes \(\frac{(x + 1)(x + 3)}{2x + 1}<0\).

Step2: Find critical points

The critical points are the values of \(x\) that make the numerator or the denominator zero.

• For the numerator: \((x + 1)(x + 3)=0\) gives \(x=-3\) or \(x=-1\).
• For the denominator: \(2x + 1 = 0\) gives \(x=-\frac{1}{2}\).

These critical points divide the real number line into four intervals: \((-\infty,-3)\), \((-3,-1)\), \((-1,-\frac{1}{2})\), and \((-\frac{1}{2},\infty)\).

Step3: Test intervals

• For \(x\in(-\infty,-3)\), let's take \(x = -4\). Then \(\frac{(-4 + 1)(-4 + 3)}{2(-4)+1}=\frac{(-3)(-1)}{-8 + 1}=\frac{3}{-7}<0\). So this interval is part of the solution.
• For \(x\in(-3,-1)\), let's take \(x=-2\). Then \(\frac{(-2 + 1)(-2 + 3)}{2(-2)+1}=\frac{(-1)(1)}{-4 + 1}=\frac{-1}{-3}=\frac{1}{3}>0\). So this interval is not part of the solution.
• For \(x\in(-1,-\frac{1}{2})\), let's take \(x = -\frac{3}{4}\). Then \(\frac{(-\frac{3}{4}+ 1)(-\frac{3}{4}+ 3)}{2(-\frac{3}{4})+1}=\frac{(\frac{1}{4})(\frac{9}{4})}{-\frac{3}{2}+1}=\frac{\frac{9}{16}}{-\frac{1}{2}}=-\frac{9}{8}<0\). So this interval is part of the solution.
• For \(x\in(-\frac{1}{2},\infty)\), let's take \(x = 0\). Then \(\frac{(0 + 1)(0 + 3)}{2(0)+1}=\frac{3}{1}=3>0\). So this interval is not part of the solution.

We also need to exclude the critical points that make the denominator zero (\(x = -\frac{1}{2}\)) and the points that make the numerator zero (but we check the sign around them). The points \(x=-3\) and \(x = -1\) make the numerator zero, and when \(x=-3\) or \(x=-1\), the left - hand side of the inequality is zero, which does not satisfy the strict inequality \(<0\), so we exclude them. The point \(x = -\frac{1}{2}\) makes the denominator zero, so it is also excluded.

Answer:

The solution to the inequality \(\frac{x^{2}+4x + 3}{2x + 1}<0\) is \(x\in(-\infty,-3)\cup(-1,-\frac{1}{2})\)